A solution of 0.20 g of a non-electrolyte in 2 g of water freezes at 271.14 K . If `K_(f) = 1.86 K "molality"^(-1) ` then the molar mass of the solute is :

Options

A. 207.8 g//mol
B. 179.79 g//mol
C. 200.8 g//mol
D. 100.0 g//mol

Answer

D

Explanation

`Delta T_(F)=(1000_(f) xx W_(2))/(M_(2) W_(1))`
or `1.86=(1000 xx 1.86 xx 0.20)/(M_(2) xx 20)`
`:. M_(2) = (1000 xx 1.86 xx 0.20)/(1.86 xx 20 )=100.0`